I have a solution using cordinates geometry! If no one find a more nice solution, i'll post!

I have a nice and short metric solution, so here it goes: We have that $AN=\frac{AQ}{\cos{QAN}}$ and $AM=\frac{AP}{\cos{PAM}}$. But $AQ=\frac{x}{\cos{QAC}}$ and $AP=\frac{x}{\cos{PAB}}$, where $x=\frac{AB}{2}$. So $AM=\frac{x}{\cos{PAM}\cdot \cos{PAB}}$ and $AN=\frac{x}{\cos{QAN}\cdot\cos{QAC}}$, so the inequality rewrites as: $\cos{QAC}\cdot \cos{QAN}+\cos{PAM}\cdot \cos{PAB}\leq 1$ $\iff \cos{B}+\cos{(QAP-\frac{A}{2})}+\cos{BAM}+\cos{(QAP-\frac{A}{2})}\leq 2$. But $\angle{B}=180-\angle{BAM}$, therefore $\cos{B}=-\cos{BAM}$, thus the inequality rewrites as $\cos{(QAP-\frac{A}{2})}\leq 1$, which is true.

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Virgil Nicula wrote: Let $ABC$ be an $A$- isosceles triangle and let $d$ be a line for which $\{\begin{array}{c}A\in d\\\ d\parallel BC\end{array}$. Given are two points $P$, $Q$ such that $\{\begin{array}{c}PA=PB\\\ QA=QC\end{array}$. Denote the points $\{M,N\}\subset d$ for which $\{\begin{array}{c}PM\perp PA\\\ QN\perp QA\end{array}$. Prove that $\frac{1}{AM}+\frac1{AN}\leq\frac2{AB}$. I think isn't necessary the condition $PQ\perp BC$. Maybe I was mistaken . The all is perhaps ! Proof. Denote $\{\begin{array}{c}m(\widehat{MAP})=x\\\ m(\widehat{NAQ})=y\end{array}$. Suppose w.l.o.g. $0\le y\le x\le B$. Thus, $\{\begin{array}{c}AB=2\cdot AP\cdot\cos \widehat{BAP}=-2\cdot AP\cdot \cos (B+x)=-2\cdot AM\cdot\cos x\cdot\cos (B+x)\\\ AC=2\cdot AQ\cdot\cos \widehat{CAQ}=2\cdot AQ\cdot\cos (B-y)=2\cdot AN\cdot\cos y\cdot\cos (B-y)\end{array}\|$ $\implies$ $\frac{1}{AM}+\frac{1}{AN}=$ $\frac{2}{AB}\cdot[\cos y\cos (B-y)-\cos x\cos (B+x)]=$ $\frac{1}{AB}\cdot [\cos (B-2y)-\cos (B+2x)]=$ $\frac{2}{AB}\cdot \sin (B+x-y)\sin (x+y)$ $\implies$ $\frac{1}{AM}+\frac{1}{AN}\le\frac{2}{AB}$, with equality iff $x=90-\frac{B}{2}$ and $y=\frac{B}{2}$. Exists at least a mistake, please ?! Remark. Denote the circumcenter $O$ of $\triangle ABC$. Thus, $PQ\perp BC\Longleftrightarrow$ $OP=OQ\Longleftrightarrow$ $AP\cos x=AQ\cos y$. But $\frac{AP}{AO}=-\frac{\sin B}{\cos (B+x)}$ and $\frac{AQ}{AO}=\frac{\sin B}{\cos (B-y)}$. Therefore, $\cos x\cos (B-y)+\cos y\cos (B+x)=0$ $\Longleftrightarrow$ $\cos (B+x-y)+\cos (x+y)\cos B=0$.

I think are you sure $x+y\le\pi$ when you get $\sin (B+x-y)\sin (x+y) \le 1$ you must have $\sin(B+x-y)\ge 0$ and $\sin(x+y)\ge 0$ or $0\le B+x-y,x+y\le\pi$

Yes, $0\le y\le x\le B\Longrightarrow$ $\{\begin{array}{c}0<B+x-y\le 2B<180\\\ 0\le x+y\le 2B<180\end{array}$.